# Ex 2.2 sets ncert maths solution class 11

# EXERCISE 2.2 (Class 11)

**1. Let A = {1, 2, 3, … , 14}. Define a relation R from A to A by R = {( x, y): 3x – y = 0, where x, y ∈ A}. Write down its domain, codomain and range.**

**Solution: **The relation R from A to A is given as:

R = {(*x*, *y*): 3*x* – *y* = 0, where *x*, *y* ∈ A}

= {(*x*, *y*): 3*x* = *y*, where *x*, *y* ∈ A}

Hence,

R = {(1, 3), (2, 6), (3, 9), (4, 12)}

Now,

Hence, Domain of R = {1, 2, 3, 4}

The whole set A is the codomain of the R.

Hence, Codomain of R = {1, 2, 3, …, 14} = A

The range of R is the set of all second elements of the ordered pairs in the relation.

Hence, Range of R = {3, 6, 9, 12}

**2. Define a relation R on the set N of natural numbers by R = {( x, y): y = x + 5, x is a natural number less than 4; x, y ∈ N}. Depict this relationship using roster form. Write down the domain and the range.**

**Solution: ****The relation R is given by:**

R = {(*x*, *y*): *y* = *x* + 5, *x* is a natural number less than 4, *x*, *y* ∈ **N**}

The natural numbers less than 4 ={1, 2, 3}

R = {(1, 6), (2, 7), (3, 8)}

Now,

The domain of R is the set of all first elements of the ordered pairs in R.

Hence, Domain of R = {1, 2, 3}

The range of R is the set of all second elements of the ordered pairs in R

Hence, Range of R = {6, 7, 8}

**3. A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by R = {( x, y): the difference between x and y is odd; x ∈ A, y ∈ B}. Write R in roster form.**

**Solution: **Given,

A = {1, 2, 3, 5} and B = {4, 6, 9}

The relation from A to B is given as

R = {(*x*, *y*): the difference between *x* and *y* is odd; *x* ∈ A, *y *∈ B}

R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}

**4. The figure shows a relationship between the sets P and Q. Write this relation**

**(i) in set-builder form (ii) in roster form**

**What is its domain and range?**

**Solution: **From the figure, it’s seen that

P = {5, 6, 7}, Q = {3, 4, 5}

The relation between P and Q:

Set-builder form

**(i)** R = {(*x, y*): *y = x* – 2; *x* ∈ P}

Roster form

**(ii)** R = {(5, 3), (6, 4), (7, 5)}

Domain of R = {5, 6, 7}

Range of R = {3, 4, 5}

**5. Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by**

**{( a, b): a, b ∈ A, b is exactly divisible by a}.**

**(i) Write R in roster form**

**(ii) Find the domain of R**

**(iii) Find the range of R**

**Solution: **Given,

A = {1, 2, 3, 4, 6} and

relation R = {(*a*, *b*): *a*, *b* ∈ A, *b* is exactly divisible by *a*}

Hence,

**(i)** R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)}

**(ii)** Domain of R = {1, 2, 3, 4, 6}

**(iii)** Range of R = {1, 2, 3, 4, 6}

**6. Determine the domain and range of the relation R defined by R = {( x, x + 5): x ∈ {0, 1, 2, 3, 4, 5}}.**

**Solution: **Given,

Relation R = {(*x*, *x* + 5): *x* ∈ {0, 1, 2, 3, 4, 5}}

R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}

So,

Domain of R = {0, 1, 2, 3, 4, 5} and,

Range of R = {5, 6, 7, 8, 9, 10}

**7. Write the relation R = {( x, x^{3}): x is a prime number less than 10} in roster form.**

**Solution: **Given,

Relation R = {(*x*, *x*^{3}): *x *is a prime number less than 10}

The prime numbers less than 10 = {2, 3, 5, 7}.

Therefore, R = {(2, 8), (3, 27), (5, 125), (7, 343)}

**8. Let A = { x, y, z} and B = {1, 2}. Find the number of relations from A to B.**

**Solution: **Given, A = {*x*, *y*, z} and B = {1, 2}

Now,

A × B = {(*x*, 1), (*x*, 2), (*y*, 1), (*y*, 2), (*z*, 1), (*z*, 2)}

As *n*(A × B) = 6,

the number of subsets of A × B = 2^{6}.

Thus, the number of relations from A to B = 2^{6}.

**9. Let R be the relation on Z defined by R = {( a, b): a, b ∈ Z, a – b is an integer}. Find the domain and range of R.**

**Solution: **Given,

Relation R = {(*a*, *b*): *a*, *b* ∈ Z, *a *– *b* is an integer}

We know that the difference between any two integers is always an integer.

Therefore,

Domain of R = Z

Range of R = Z.